Answer:
Cu = 1453.72J/Kg°C
The heat capacity of the liquid is 1453.72J/Kg°C
Explanation:
At equilibrium, assuming no heat loss to the surrounding we can say that;
Heat gained by ice + heat gained by cold water = heat loss by hot unknown liquid (24°C) + heat loss by aluminium calorimeter.
Given;
Mass of ice = mass of cold water = mc = 30g = 0.03kg
Mass of hot unknown liquid mh= 300g = 0.3kg
Mass of aluminium calorimeter ma= 100g = 0.1 kg
change in temperature cold ∆Tc = (4-0) = 4°C
Change in temperature hot ∆Th = 24-4 = 20°C
Specific heat capacity of water Cw= 4186J/Kg°C
Specific heat capacity of aluminium Ca = 900J/kg°C
Specific heat capacity of unknown liquid Cu =?
Heat of condensation of ice Li = 334000J/Kg
So, the statement above can be written as.
mcLi + mcCw∆Tc = maCa∆Th + mhCu∆Th
Making Cu the subject of formula, we have;
Cu = [mcLi + mcCw∆Tc - maCa∆Th]/mh∆Th
Substituting the values we have;
Cu = (0.03×334000 + 0.03×4186×4 - 0.1×900×20)/(0.3×20)
Cu = 1453.72J/Kg°C
the heat capacity of the liquid is 1453.72J/Kg°C
If the volume of one drop is 0.031 mL according to Stu Dent’s measurement, approximately what volume would 22 drops be? Answer with two significant digits and units of mL.
Answer:
Volume of 22 drop will be 0.68 ml
Explanation:
We have given volume of one drop = 0.031 ml
We know that 1 liter = 1000 ml
So [tex]1ml=10^{-3}L[/tex]
So 0.031 ml will be equal to [tex]0.031\times 10^{-3}L[/tex]
We have to find the volume of 22 drop
For finding volume of 22 drop we have to multiply volume of one drop by 22
So volume of 22 drop will be [tex]=22\times 0.031\times 10^{-3}=0.682\times 10^{-3}L=0.68mL[/tex]
So volume of 22 drop will be 0.68 ml
Two vectors, and , lie in the xy plane. Their magnitudes are 4.25 and 7.45 units, respectively, and their directions are 312° and 86.0°, respectively, as measured counterclockwise from the positive x axis. What are the values of the following products?
A) (r)(s)= -18.3
B) (r)X(s)=?
Answer:
A) r•s = -21.99 units
B) rXs = 22.77 units (In the direction of k)
Completed question
Two vectors, r and s, lie in the xy plane. Their magnitudes are 4.25 and 7.45 units, respectively, and their directions are 312° and 86.0°, respectively, as measured counterclockwise from the positive x axis. What are the values of the following products?
A) r•s =
B) (r)X(s)=
Explanation:
A) dot product of r and s can be expressed mathematically as;
r•s = |r||s|cos(A) .......1
Where A is the angle between the two vectors.
|r| and |s| are the magnitude of vector r and s.
r•s = 4.25×7.45 ×cos(86-312) = -21.99 units
B) cross product of r and s can be expressed mathematically as;
rXs = |r||s|sin(A) .......2
Where A is the smallest angle between the two vectors
|r| and |s| are the magnitude of vector r and s.
A = 312-86 = 226
A = 360-226 = 134 smallest
rXs = 4.25 × 7.45 × sin134.
rXs = 22.77 units
In the direction of k
(a) -22.00 units
(b) 22.77 units
Explanation:Given vectors are r and s
Where;
r = |r| = 4.25 and ∠r = 312° measured anticlockwise
s = |s| = 7.45 and ∠s = 86° measured anticlockwise
First, let's calculate the angle between vectors r and s by representing them in the figure below;
y | /s
| /
| /
| /
|/ )86° x
|\) 48°
| \
| \
| \
| \ r
To get the acute angle between r and the +x axis, subtract the reflex angle of r (312°) from 360° as follows;
360 - 312 = 48°
As shown in the diagram, the angle between vectors r and s is 48° + 86° = 134°
Now,
(a) The (r)(s) represents the dot or scalar product of the two vectors and it is given as;
(r) (s) = r x s cos θ ---------------------------(i)
Where;
r = magnitude of vector r = 4.25
s = magnitude of vector s = 7.45
θ is the angle between the two vectors r and s = 134°
Substitute these values into equation (i) as follows;
(r) (s) = 4.25 x 7.45 cos 134°
(r) (s) = 4.25 x 7.45 x -0.6947
(r) (s) = 31.66 x -0.6947
(r) (s) = -22.00 units
(b) The (r) X (s) represents the vector product of the two vectors and it is given as;
(r) (s) = r x s sin θ ---------------------------(ii)
Where;
r = magnitude of vector r = 4.25
s = magnitude of vector s = 7.45
θ is the angle between the two vectors r and s = 134°
Substitute these values into equation (ii) as follows;
(r) (s) = 4.25 x 7.45 sin 134°
(r) (s) = 4.25 x 7.45 x 0.7193
(r) (s) = 31.66 x 0.7193
(r) (s) = 22.77 units
A 5 kg mass is attached to the ceiling of an elevator by a rope whose mass is negligible. What is the tension in the rope when the elevator accelerates upward at 4 m/s2?
Answer:
Explanation:
Given
mass of object [tex]m=5\ kg[/tex]
Object is attached to the ceiling of an elevator by a rope
Suppose T is the Tension in the rope so
acceleration of the elevator [tex]a=4\ m/s^2[/tex] (upward)
From Free Body diagram we can write as
[tex]T-mg=ma[/tex]
[tex]T=m(g+a)[/tex]
[tex]T=5(9.8+4)[/tex]
[tex]T=69\ N[/tex]
You and your best friend are trying to pull one another toward your respective dorm rooms. You're the stronger of the two and, with a mighty tug, you drag your friend into your room.
As you are pulling your friend toward your room, the force you exert on your friend is:
A) equal in amount to the force your friend exerts on you.
B) definitely equal to three times the weight of Spongebob Squarepants.
C) less in amount than the force your friend exerts on you.
D) greater in amount than the force your friend exerts on you.
Answer:
A) equal in amount to the force your friend exerts on you.
Explanation:
According to the Newton's third law of motion every action has equal and opposite reaction. Here both the bodies exert equal and opposite forces on each other but it is just that I am able to stop myself from getting pulled by a greater force of friction by applying more normal force on the ground.
As we know that the force of friction is given as:
[tex]f=\mu.N[/tex]
where:
[tex]\rm f=\ force\ of\ friction\\N=\ applied\ normal\ force\ to\ the\ ground\\ \mu=\ coefficient\ of\ friction[/tex]
A battery having an emf of 9.63 V delivers 118 mA when connected to a 60.0 Ω load. Determine the internal resistance of the battery.
Answer:
21.6 ohm
Explanation:
We are given that
EMF=E=9.63 V
Current=I=118 mA=[tex]118\times 10^{-3} A[/tex]
[tex]1 mA=10^{-3} A[/tex]
Resistance=[tex]R=60\Omega[/tex]
We have to find the internal resistance of the battery.
We know that
[tex]V=E-Ir[/tex]
We know that V=IR
[tex]IR=E-Ir[/tex]
[tex]IR+Ir=E[/tex]
[tex]I(R+r)=E[/tex]
[tex]R+r=\frac{E}{I}[/tex]
Substitute the values
[tex]60+r=\frac{9.63}{118\times 10^{-3}}[/tex]
[tex]60+r=81.6[/tex]
[tex]r=81.6-60[/tex]
[tex]r=21.6\Omega[/tex]
Hence, the internal resistance of the battery=21.6 ohm
Answer:
21.6 ohm
Explanation:
EMF of the battery, E = 9.63 V
Current, i = 118 mA = 0.118 A
Resistance, R = 60 ohm
Let the internal resistance of the cell is r.
[tex]i = \frac{E}{R + r}[/tex]
R + r = 9.63 / 0.118
60 + r = 81.6
r = 21.6 ohm
An Atwood machine is constructed using a disk of mass 2 kg and radius 24.8 cm. What is the acceleration of the system? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s 2 .
Answer:
a = (m₂-m₁) / (m₂ + m₁ + ½ m)
a = (m₂-m₁) / (m₂ + m₁ + 1)
Explanation:
An Atwood machine consists of two masses of different m1 and m2 value that pass through a pulley, in this case with mass. Let's use Newton's second law for this problem.
Assume that m₂> m₁, so the direction of descent of m₂ is positive, this implies that the direction of ascent of m₁ is positive
Equation of the side of m₁
T₁ - W₁ = m₁ a
Equation of the side of m₂
W₂ - T₁ = m₂ a
Mass pulley equation m; by convention the counterclockwise rotation is positive
τ = I α
T₁ R - T₂ R = I (-α)
The moment of inertia of a disk is
I = ½ m R²
Angular and linear acceleration are related
a = α R
α = a / R
The rotation is clockwise, so it is negative
We replace
(T₁ –T₂) R = ½ m R² (-a / R)
T₁ -T₂ = - ½ m a
Let's write our three equations together
T₁ - m₁ g = m₁ a
m₂ g - T₂ = m₂ a
T₁ –T₂ = -½ m a
Let's multiply the last equation by (-1) and add
m₂ g - m₁ g = m₂ a + m₁ a + ½ m a
a = (m₂-m₁) / (m₂ + m₁ + ½ m)
calculate
a = (m₂ - m₁)/ (m₁ +m₂ + 1)
Based on the calculations, the acceleration of this system is equal to 0.58 [tex]m/s^2[/tex].
Given the following data:
Mass of disk = 2 kg.Radius of disk = 24.8 cm.Acceleration of gravity = 9.8 [tex]m/s^2[/tex]How to calculate the acceleration of the system.First of all, we would determine the moment of inertia of this disk by using this formula:
[tex]I=\frac{1}{2} mr^2\\\\I=\frac{1}{2} \times 2 \times (0.248)^2\\\\I=0.0615\;Kgm^2[/tex]
Next, we would use a free body diagram to determine the tensional forces acting on the disk by applying Newton's Second Law of Motion as follows:
For the first force:
[tex]F_1g-F_{T1}=m_1a\\\\m_1g-F_{T1}=m_1a\\\\1.61(9.8)-F_{T1}=1.61a\\\\15.8-F_{T1}=1.61a\\\\F_{T1}=15.8-1.61a[/tex]
For the second force:
[tex]F_{T2}-F_2g=m_2a\\\\F_{T2}-m_2g=m_2a\\\\F_{T2}-1.38(9.8)=1.38a\\\\F_{T2}-13.5=1.38a\\\\F_{T2}=13.5+1.38a[/tex]
For the torque, we have:
[tex]\sum T =I\alpha \\\\F_{T1}r-F_{T2}r=I\alpha\\\\(F_{T1}-F_{T2})r=I\alpha\\\\(15.8-1.61a-[13.5+1.38a])0.248=0.0615 \times \frac{a}{0.248} \\\\(15.8-1.61a-13.5-1.38a)0.248=0.0615 \times \frac{a}{0.248} \\\\(2.3-2.99a)0.248=0.0615 \times \frac{a}{0.248}\\\\0.5704-0.7415a=\frac{0.0615a}{0.248}\\\\0.1415-0.1839a=0.0615a\\\\0.1839a+0.0615a=0.1415\\\\0.2454a=0.1415\\\\a=\frac{0.1415}{0.2454}[/tex]
Acceleration, a = 0.58 [tex]m/s^2[/tex]
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A man wandering in the desert walks 2.7 miles in the direction S 35°W. He then turns 90° and walks 3.5 miles in the direction N 55° W. At that time, how far is he from his starting point, and what is his bearing from his starting point?
To solve this problem we will make a diagram in the Cartesian plane that will allow us to find and understand more accurately the displacement and the angle of rotation.
According to Pythagoras, the distance traveled would be equivalent to
[tex]d = \sqrt{(2.7)^2+(3.5)^2}[/tex]
[tex]d = 4.4 miles[/tex]
The individual had a displacement of 4.4 thousand from the starting point.
Now the angle [tex]\theta[/tex] plus the previously given angle will allow us to find the direction of travel.
[tex]tan\theta = \frac{\text{Opposite side}}{\text{Adjacent side}}[/tex]
[tex]tan\theta = \frac{3.5}{2.7}[/tex]
[tex]\theta = tan^{-1} (\frac{3.5}{2.7})[/tex]
[tex]\theta = 52.35\°[/tex]
[tex]\angle =[/tex] [tex]\theta + 35 = 52.35+35 = 87.35\°[/tex]
Therefore the net direction of the man is S 87.35° W
A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between his clothes and Earth is 0.635. He slides so that his speed is zero just as he reaches the base. The acceleration of gravity is 9.8 m/s 2 . What is the magnitude of the mechanical energy lost due to friction acting on the runner? Answer in units of J.
To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:
[tex]\Delta W = \Delta KE[/tex]
[tex]\Delta W = \frac{1}{2} mv^2[/tex]
Here,
m = mass
v = Velocity
Our values are given as,
[tex]m = 79.7kg[/tex]
[tex]v = 4.77m/s[/tex]
Replacing,
[tex]\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2[/tex]
[tex]\Delta W = 907J[/tex]
Therefore the mechanical energy lost due to friction acting on the runner is 907J
You are red and your friend is green. You stand 2 meters from the mirror. Your friend stands 1 meter from the mirror. Would your friend appear to be in a different position to anyone else, in a different position?
Answer:
Explanation:
It is given that red is 2 m from the mirror and green is 1 m from the mirror so the image of green and red will be formed 1 and 2 m behind the mirror respectively.
Green will be seen at the same distance from the mirror when seen from different position to anyone else.
The above can be explained by the given diagram
A 2.0 m × 4.0 m horizontal plastic sheet has a charge of −10 μC , uniformly distributed. A tiny 4.0 μg plastic sphere is suspended motionless just above the center of the sheet. What is the charge on the sphere?
Answer:
Explanation:
Given
Dimension of Plastic sheet is [tex]2\times 4\ m^2[/tex]
Charge on sheet [tex]Q=-10\ \mu C[/tex]
Charge density [tex]\sigma =\frac{q}{A}[/tex]
[tex]\sigma =\frac{-10\times 10^{-6}}{8}=1.25\times 10^{-6}\ C/m^2[/tex]
Sphere has mass of [tex]m=4\ \mug[/tex]
If sphere is suspended motionless then its weight is balanced by repulsion force
Repulsive force [tex]F_r=qE[/tex]
where E=Electric field due to sheet
[tex]E=\frac{q}{2\epsilon _0}[/tex]
[tex]E=\frac{-10\times 10^{-6}}{2\times 8.85\times 10^{-12}}[/tex]
[tex]E=5.647\times 10^{5}\ N/C[/tex]
[tex]F_r=q\times 5.647\times 10^{5}[/tex]
[tex]F_r=mg[/tex]
[tex]q=\frac{mg}{E}[/tex]
[tex]q=\frac{4\times 10^{-6}\times 9.8}{5.647\times 10^{5}}[/tex]
[tex]q=6.9417\times 10^{-11}\ C[/tex]
A 3.0 L cylinder is heated from an initial temperature of 273 K at a pressure of 105 kPa to a final temperature of 367 K. Assuming the amount of gas and the volume remain the same, what is the pressure (in kilopascals) of the cylinder after being heated?
Answer:
P₂= 141.15 kPa
Explanation:
Given that
Volume ,V= 3 L
V= 0.003 m³
Initial temperature ,T₁ = 273 K
Initial pressure ,P₁ = 105 kPa
Final temperature ,T₂ = 367 K
Given that volume of the cylinder is constant .
Lets take final pressure = P₂
We know that for constant volume process
[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}\\P_2=P_1\times \dfrac{T_2}{T_1}\\P_2=105\times \dfrac{367}{273}\ kPa\\\\P_2=141.15\ kPa[/tex]
Therefore the final pressure = 141.15 kPa
P₂= 141.15 kPa
Which of the following does not involve work? 1. A child is pushed on a swing. 2. A golf ball is struck. 3. A weight lifter does military presses (lifting weights over his head.) 4. A professor picks up a piece of chalk from the floor. 5. A runner stretches by pushing against a wall.
From the work theorem, this is defined as the amount of force applied on an object displaced on a longitudinal unit. Mathematically this is
[tex]W = \vec{F} \times \vec{d}[/tex]
Here,
F = Force vector
d = Displacement vector
Of all the options presented, only in the last one there is no change in distance, so the work done there is zero.
The correct option is 5.
Work in Physics is when a force causes displacement. Of the choices, the example of a runner stretching by pushing against a wall does not involve work, as there is no displacement.
Explanation:In the context of Physics, 'work' is defined as a force causing displacement on an object. Essentially, work is done when a force acts upon an object to cause or prevent motion. Among the options provided, 5. A runner stretches by pushing against a wall does not involve 'work'. This is because, despite the runner exerting a force against the wall, there is no displacement of the wall in response to this force.
Work (W) is calculated by multiplying the force (F) that is applied to an object and the distance (d) that the object is moved, i.e., W = F * d. In this case, since the displacement (d) is zero, the work done is also zero.
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What is the boiling point of an aqueous solution that freezes at -2.05 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your answer using 2 decimal places!!!!
To solve this problem we will apply the concepts of Boiling Point elevation and Freezing Point Depression. The mathematical expression that allows us to find the temperature range in which these phenomena occur is given by
Boiling Point Elevation
[tex]\Delta T_b = K_b m[/tex]
Here,
[tex]K_b[/tex] = Constant ( Different for each solvent)
m = Molality
Freezing Point Depression
[tex]\Delta T_f = K_f m[/tex]
Here,
[tex]K_f =[/tex] Constant ( Different for each solvent)
m = Molality
According to the statement we have that
[tex]\Delta T_f = T_0 -T_f = 0-(-2.05)[/tex]
[tex]\Delta T_f = 2.05\°C[/tex]
From the two previous relation we can find the ratio between them, therefore
[tex]\frac{\Delta T_b}{\Delta T_f} = \frac{K_b}{K_f}[/tex]
[tex]\frac{\Delta T_b}{\Delta T_f} = \frac{0.512}{1.86}[/tex]
[tex]\frac{\Delta T_b}{\Delta T_f} = 0.275[/tex]
We already know the change in the freezing point, then
[tex]\Delta T_b = 0.275 (\Delta T_f)[/tex]
[tex]\Delta T_b = 0.275 (2.05)[/tex]
[tex]\Delta T_b = 0.5643\°C[/tex]
The temperature difference in the boiling point is 100°C (Aqueous solution), therefore
[tex]T_b -100 = 0.5643[/tex]
[tex]T_b = 100.56\°C[/tex]
Therefore the boiling point of an aqueous solution is [tex]100.56\°C[/tex]
Final answer:
The boiling point of the aqueous solution is 99.44 °C.
Explanation:
The boiling point of an aqueous solution can be determined using the formula:
Tbp = Tbpsolvent + (Kbp * m)
where:
Tbp is the boiling point of the solution
Tbpsolvent is the boiling point of the pure solvent
Kbp is the ebullioscopic constant for the solvent
m is the molality of the solution
In this case, we are given that the freezing point of the solution is -2.05 degrees C. We can use the formula for freezing point depression to find the molality:
AT = Kf * m
Rearranging the formula, we can solve for m:
m = AT / Kf
Substituting the given values:
m = (-2.05 °C) / (1.86 °C kg.mol-¹)
Rounding to 2 decimal places:
m = -1.10 mol/kg
Now, we can use the formula for boiling point elevation to find the boiling point:
Tbp = Tbpsolvent + (Kbp * m)
Substituting the given values:
Tbp = 100.00 °C + (0.512 °C/m * -1.10 mol/kg)
Calculating:
Tbp = 100.00 °C - 0.56 °C = 99.44 °C
Therefore, the boiling point of the aqueous solution is 99.44 °C.
A hydraulic turbine is used to generate power by using the water in a dam. The elevation difference between the free surfaces upstream and downstream of the dam is 131 m. The water is supplied to the turbine at a rate of 201 kg/s. If the shaft power output from the turbine is 234 kW, the efficiency of the turbine is_________.
Answer:
0.906
Explanation:
Let g = 9.81 m/s2. We can calculate the rate of change in potential energy when m = 201kg of water is falling down a distance of h = 131m per second
[tex]\dot{E_p} = \dot{m}gh = 201*9.81*131 = 258307 J/s (W) = 258.307 kW[/tex]
So the efficiency of the water turbine is the ratio of output power over input power:
[tex]\frac{234}{258.307} = 0.906[/tex]
This question involves the concepts of efficiency and potential energy.
The efficiency of the turbine is "90.6%".
The efficiency of the turbine can be given by the following formula:
[tex]Efficiency = \frac{Turbine\ shaft\ power}{Potential\ Energy\ power\ of\ Water}\\\\[/tex]
where,
Turbine shaft power = 234 KW
Potential energy power of water = mgh/t = (201 kg/s)(9.81 m/s²)(131 m)
Potential energy power of water = 258307.11 W = 258.31 KW
Therefore,
[tex]Efficiency=\frac{234\ KW}{258.31\ KW}[/tex]
Efficiency = 0.906 = 90.6%
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A 15-turn circular wire loop with a radius of 3.0 cm is initially in a uniform magnetic field with a strength of 0.5 T. The field decreases to zero over a time of 0.10 sec. What is the induced emf?
Answer:
0.212V
Explanation:
The induced emf in circular loop is [tex]\epsilon=-d\frac{\phi_B}{dt}=-NA\frac{dB}{dt}[/tex]
N = Number of loops = 15
A = Cross sectional Area = [tex]\pi[/tex][tex]0.03^{2}[/tex] = 0.0009[tex]\pi[/tex]
dB = change in magnetic Field = 0-0.5 = -0.5
dt = time taken = 0.1sec
[tex]\epsilon=-15*0.0009\pi *\frac{-0.5}{0.1}[/tex] = 0.212V
An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.50 m away at an average speed of 3.00 m/s, returning just in time to catch the falling ball. (a) With what minimum initial speed must she throw the ball upward to accomplish this feat? (b) How high above its initial position is the ball just as she reaches the table?
To accomplish this feat, the entertainer must throw the ball upward with a minimum initial speed and reach a certain height above its initial position.
Explanation:(a) While the ball is in the air, it rises and then falls to a final position 10.0 m higher than its starting altitude. We can find the time for this by using Equation 4.22:
y = yo + voyt - (1/2)gt².
If we take the initial position yo to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity:
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Final answer:
The minimum initial speed the entertainer must throw the ball upward is 17.96 m/s, and the height of the ball just as she reaches the table is approximately 15.56 m.
Explanation:
Calculating the Minimum Initial Speed and Height of the Juggling Ball
To answer the student's question regarding the juggling entertainer's act, we need to apply concepts from kinematics, a subfield of classical mechanics in physics. First, let's find out the total time the entertainer has to throw the ball and return to the starting position. We can find the time taken to reach the table and come back with the given average speed and distance: Time = (2 × Distance) / Speed. The entertainer runs to and from a table which is 5.50 m away at an average speed of 3.00 m/s. Hence, the total time for the round trip is (2 × 5.50 m) / 3.00 m/s = 3.67 s.
Now, we use the equation of motion to calculate the initial vertical speed needed for the ball to be in the air for this duration: s = ut + 0.5 × a × t², where s is the displacement (which is 0 because the ball returns to the same position), u is the initial vertical speed, and a is the acceleration due to gravity (-9.81 m/s²). After rearranging the formulas, the initial speed u turns out to be 17.96 m/s.
To find the height of the ball as she reaches the table at 5.50 m away, we consider half the total time, which is 1.835 s. Putting this in the kinematic equation s = ut + 0.5 × a × t², we find the height to be approximately 15.56 m at that instant.
A gas has a pressure of 48atm in a 15.5L container. It was found that at 25∘C the gas occupied a volume of 25L and had a pressure of 22atm. What was the initial temperature in degrees Celsius?
Answer:
130.165636364°C
Explanation:
P = Pressure
V = Volume
n = Number of moles
R = Gas constant = 0.082 L atm/mol K
From ideal gas law we have
[tex]PV=nRT\\\Rightarrow n=\dfrac{PV}{RT}\\\Rightarrow n=\dfrac{22\times 25}{0.082\times (25+273.15)}\\\Rightarrow n=22.496451696\ moles[/tex]
[tex]PV=nRT\\\Rightarrow T=\dfrac{PV}{nR}\\\Rightarrow T=\dfrac{48\times 15.5}{22.496451696\times 0.082}\\\Rightarrow T=403.315636364\ K[/tex]
The initial temperature is [tex]403.315636364-273.15=130.165636364\ ^{\circ}C[/tex]
Answer: 130 degrees Celsius
Explanation:
P1V1 / T1 = P2V2 / T2
Let the subscript 1 represent the initial 15.5L of gas and the subscript 2 represent the gas at the final volume of 25L. Rewrite the temperature in Kelvin by adding 273.
P1=48atm, V1=15.5L, P2=22atm, V2=25L, T2=298K, and T1 is unknown.
Substitute the known values into the combined gas law equation.
P1V1 / T1=P2V2 / T2 → (48atm)(15.5L) / T1 = (22atm)(25L) / 298K
Solve the equation for T1, simplify, and round to the nearest degree.
T1 = P1V1T2 / P2V2
T1 = (48atm)(15.5L)(298K)(22atm)(25L)
T1 = 403K
To get the temperature in Celsius, subtract 273.
T1 = 130∘C
Suppose the average mass of each of 20,000 asteroids in the solar system is 1017 kg. Compare the total mass of these asteroids to the mass of Earth. Assuming a spherical shape and a density of 3000 kg/m3, estimate the diameter of an asteroid having this average mass.
Answer:
The mass of the asteroids is 0.000334896182184 times the mass of the Earth.
39929.4542466 m
Explanation:
Total mass of the asteroids
[tex]m_a20000\times 10^{17}=2\times 10^{21}\ kg[/tex]
[tex]m_e[/tex] = Mass of Earth = [tex]5.972\times 10^{24}\ kg[/tex]
The ratio is
[tex]\dfrac{m_a}{m_e}=\dfrac{2\times 10^{21}}{5.972\times 10^{24}}\\\Rightarrow \dfrac{m_a}{m_e}=0.000334896182184[/tex]
The mass of the asteroids is 0.000334896182184 times the mass of the Earth.
Volume is given by
[tex]V=\dfrac{m}{\rho}\\\Rightarrow \dfrac{4\pi}{3\times 8} d^3=\dfrac{m}{\rho}\\\Rightarrow d^3=\dfrac{3\times 8}{4\pi}\dfrac{m}{\rho}\\\Rightarrow d=(\dfrac{3\times 8}{4\pi}\dfrac{m}{\rho})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{3\times 8}{4\pi}\dfrac{10^{17}}{3000})^{\dfrac{1}{3}}\\\Rightarrow d=39929.4542466\ m[/tex]
The diameter is 39929.4542466 m
If a certain silver wire has a resistance of 8.60 Ω at 29.0°C, what resistance will it have at 42.0°C?
Answer:
9.027 Ω
Explanation:
Using,
R = R₀(1+αΔt)
R = R₀(1+α[t₂-t₁]).......................... Equation 1
Where R = the value of the resistance at the final temperature, R₀ = the value of the resistance at the initial temperature, α = Temperature coefficient of resistance, t₂ = Final temperature, t₁ = Initial temperature.
Given: R₀ = 8.6 Ω, t₂ = 42 °C, t₁ = 29 °C
Constant : 0.003819/°C
Substitute into equation 1
R = 8.6(1+0.003819[42-29])
R = 9.027 Ω.
Hence the resistance = 9.027 Ω
9.02Ω
Explanation:The resistivity of a conductor increases as temperature increases. In this case, silver, which is a great conductor, will increase its resistivity linearly over a range of increasing temperatures . The relationship between resistance(R) and temperature(T) is given as;
R = R₀ (1 + α(T - T₀)) --------------(i)
where;
R and R₀ are the final and initial resistances of the material (silver in this case)
α = temperature coefficient of resistivity of the material (silver) = 0.0038/°C
T and T₀ are the final and initial temperatures.
From the question;
R₀ = 8.60Ω
T₀ = 29.0°C
T = 42.0°C
Substitute these values into equation (i);
R = 8.60 (1 + 0.0038(42.0 - 29.0))
R = 8.60(1 + 0.0038(13))
R = 8.60(1 + 0.0494)
R = 8.60(1.0494)
R = 9.02Ω
Therefore, the resistance at 42.0°C is 9.02Ω
In what regions of the electromagnetic spectrum is the atmosphere transparent enough to allow observations from the ground?
Answer:
Visible Light and Radio waves
Explanation:
The earth's atmosphere is transparent to a few windows in the electromagnetic spectrum. it is completely transparent to allow observation from the ground in visible light rang 380 to 740 nano meters. Also in the range of radio wave as communication are done from space to ground in the form of radio waves.
it is Partially transparent to Microwave and infrared range.
The atmosphere is transparent to certain regions of the electromagnetic spectrum, such as visible light, radio waves, and parts of the infrared and microwave spectra.
The Earth's atmosphere is transparent to certain regions of the electromagnetic spectrum, allowing observations to be made from the ground. These regions include the visible light spectrum, the radio waves spectrum, and parts of the infrared and microwave spectra.
In these regions, electromagnetic waves can pass through the Earth's atmosphere relatively easily, allowing ground-based observations to be conducted.
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A cannon ball is shot straight upward with a velocity of 72.50 m/s. How high is the cannon ball above the ground 3.30 seconds after it is fired? (Neglect air resistance.)
Answer:
Explanation:
Given
Cannon is fired with a velocity of [tex]u=72.50\ m/s[/tex]
Using Equation of motion
[tex]y=ut+\frac{1}{2}at^2[/tex]
where
[tex]y=displacement[/tex]
[tex]u=initial\ velocity[/tex]
[tex]a=acceleration[/tex]
[tex]t=time[/tex]
after time [tex]t=3.3 s[/tex]
[tex]y=72.50\times 3.3-\frac{1}{2}\times 9.8\times (3.3)^2[/tex]
[tex]y=239.25-53.36[/tex]
[tex]y=185.89\ m[/tex]
So after 3.3 s cannon ball is at a height of 185.89 m
The temperature of a sample of silver increased by 24.0 °C when 269 J of heat was applied. What is the mass of the sample?
Answer:
Mass of the silver will be equal to 46.70 gram
Explanation:
We have given heat required to raise the temperature of silver by 24°C is 269 J , so [tex]\Delta T=24^{\circ}C[/tex]
Specific heat of silver = 0.240 J/gram°C
We have to find the mass of silver
We know that heat required is given by
[tex]Q=mc\Delta T[/tex], here m is mass, c is specific heat of silver and [tex]\Delta T[/tex] is rise in temperature
So [tex]269=m\times 0.240\times 24[/tex]
m = 46.70 gram
So mass of the silver will be equal to 46.70 gram
A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 20.0 m above the river, whereas the opposite side is a mere 2.1 m above the river. The river itself is a raging torrent 61.0 m wide.
A) How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side?
B) What is the speed of the car just before it lands safely on the other side?
Answer:
A) The car should be traveling at 31.9 m/s.
B) The speed of the car just before it lands on the other side is 37.0 m/s.
Explanation:
Hi there!
A) Please see the attached figure for a better description of the problem. When the car reaches the other side of the river, its position vector will be r1 in the figure. The components of this vector are r1x and r1y.
If we place the origin of the frame of reference at the edge of the cliff, the components of the vector r1 will be:
r1x = 61.0 m
r1y = -20.0 m + 2.1 m = -17.9 m
The equations for the x and y-components of the position vector of the car are the following:
x = x0 + v0 · t
y = y0 + 1/2 · g · t²
Where:
x = horizontal position at a time t.
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
y = vertical position at a time t.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
Using the equation of the y-component of r1, we can find the time it takes the car to reach the other side of the river. We have to find the time at which the vector r1y is -17.9 m:
y = y0 + 1/2 · g · t² (y0 = 0 because the origin of the frame of reference is located at the edge of the cliff).
y = 1/2 · g · t²
-17.9 m = -1/2 · 9.8 m/s² · t²
-17.9 m / -4.9 m/s² = t²
t = 1.91 s
Now, using the equation of the x-component, we can find the initial velocity. We know that at t = 1.91 s, the horizontal component of the vector r1 is 61.0 m:
x = x0 + v0 · t (x0 = 0 because the origin of the frame of reference is located at the edge of the cliff).
x = v0 · t
61.0 m = v0 · 1.91 s
v0 = 61.0 m / 1.91 s = 31.9 m/s
The car should be traveling at 31.9 m/s.
B) The equation of the velocity vector of the car is the following:
v = (v0, g · t)
The horizontal component of the velocity vector is v0, 31.9 m/s.
Let's calculate the value of the vertical component:
vy = g · t
vy = -9.8 m/s² · 1.91 s
vy = -18.7 m/s
Then, the velocity vector of the car just before it lands on the other side is the following:
v = (31.9, -18.7) m/s
The magnitude of this vector is calculated as follows:
|v| = √[(31.9 m/s)² + (-18.7 m/s)²]
|v| = 37.0 m/s
The speed of the car just before it lands on the other side is 37.0 m/s.
Final answer:
To safely clear the river, the car must travel at 23.4 m/s at the cliff's edge, and it will land with a speed of 22.6 m/s on the other side.
Explanation:
A) To clear the river and land safely on the other side, the car should be traveling at a speed of 23.4 m/s as it leaves the cliff. This is calculated using the principles of projectile motion and conservation of energy.
B) The speed of the car just before it lands safely on the opposite side would be 22.6 m/s. This speed is also determined by energy considerations, such as the conversion of potential energy to kinetic energy.
the separation between the plates is now increased to 4.50 mmmm . How much energy is stored in the capacitor now?
Complete question:
A parallel-plate capacitor has plates with an area of 405 cm² and an air-filled gap between the plates that is 2.25 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery. (a) How much energy is stored in the capacitor? (b) The separation between the plates is now increased to 4.50 mm. How much energy is stored in the capacitor now?
Answer:
The energy stored in the capacitor when the plates is increased to 4.50 mm is 1.32 X 10⁻⁵ J
Explanation:
Given:
Area of the plates = 405 cm² = 405 X (10⁻²)² m²= 405 X 10⁻⁴m² = 0.0405m²
Energy stored in a capacitor = CV²/2
Where;
V is the voltage across the plates = 575 V
C is the capacitor =?
C = Kε(A/d)
K is constant = 1.0
ε is permittivity of free space = 8.885 X 10⁻¹²
d is the diameter of the two plates = 2.25 mm = 0.00225m
C = 1.0 x 8.885 X 10⁻¹² x (0.0405/0.00225)
C = 1.5993 X 10⁻¹⁰ F
(a) Energy stored in a capacitor = 0.5 X 1.5993 X 10⁻⁹ X 575²
= 2.64 X 10⁻⁵ J
(b) The separation between the plates is now increased to 4.50 mm.
C = Kε(A/d)
New diameter, d = 4.5 mm = 0.0045 m
C = 1.0 x 8.885 X 10⁻¹² x (0.0405/0.0045)
C = 7.9965 X 10 ⁻¹¹ F
Energy stored in a capacitor = 0.5 X 7.9965 X 10 ⁻¹¹ X 575²
= 1.32 X 10⁻⁵ J
Therefore, the energy stored in the capacitor when the plates is increased to 4.50 mm is 1.32 X 10⁻⁵ J
In the infinitesimal neighborhood surrounding a point in an inviscid flow, the small change in pressure, dP, that corresponds to a small change in velocity, dV, is given by the differential relation: dP=−rhoVdV. (a) Using this relation, derive a differential relation for the fractional change in density, drho/rho, as a function of the fractional change in velocity, dV/V, with the compressibility τ as a coefficient. (b) The velocity at a point in an isentropic flow of air is 10 m/s, and the density and pressure are 1.23 kg/m3 and 1.01 x 105 N/m2, respectively. The fractional change in velocity at the point is 0.01. Calculate the fractional change in density. (c) Repeat part (b), except for a local velocity at the point of 1000 m/s. Compare this result with that from part (b), and comment on the differences.
Answer:
(a). differential relation becomes dρ/ρ = -τρV2 dV/V
(b). fraction change in density; dρ/ρ = -8.7 ˣ 10⁻⁶
(c). dρ/ρ = -8.7 ˣ 10⁻²
Explanation:
Let us begin,
(a). given from the question we have that dp = -ρVdV
where dρ = ρ τ dp, i.e.
dp = dρ/ρτ ...............(1)
replacing value of dp we have,
-ρVdV = dρ/ρτ
so that dρ = -τp2 VdV
finally, dρ/ρ = -τp V2 dV/V
(b). from the question here, we were given Velocity to be = 10 m/s
density (ρ) = 1.23 kg/m3
pressure (p) = 1.01 x 10⁵ N/m2
from formula,
dρ/ρ = τs ρ V2 dV/V .............(2)
but τs = 1/γp = 1/(1.4× 1.01×10⁵) = 7.07 ˣ 10⁻⁶ m²/N
substituting value of τs into equation (2) we have
dρ/ρ = τs ρ V2 dV/V = (7.07 ˣ 10⁻⁶) ˣ (1.23) ˣ (10ˣ2) (0.01) = -8.7 ˣ 10⁻⁶
dρ/ρ = -8.7 ˣ 10⁻⁶
(c). from we question we know that dρ/ρ has a large ratio of (1000/10)²
so dρ/ρ = -8.7 ˣ 10⁻⁶ × (1000/10)² = -8.7 ˣ 10⁻²
dρ/ρ = -8.7 ˣ 10⁻².
comparing this result with part (b). we can see that when we increase the velocity of a factor 100, there is an increased factorial change in the density by a factor 104.
A particle moves so that its position (in meters) as a function of time (in seconds) is r = i ^ + 4t2 j ^ + tk ^. Write expressions for (a) its velocity and (b) its acceleration as functions of time.
Answer:
a.V=8tj+k
b.a=8j
Explanation:
Given:
Position r= i+4t^2j +tk
Nb r is position in metre and time in seconds
a.velocity is change in position/ change in time
v= ∆r/∆t =dr/dt
V=d ( i+ 4t^2j+tk)/dr
Differenting with respect to (t)
V=8tj+K
b.acceleration = change in velocity/change in time
a= ∆v/∆r =dv/dt
a=d (8tj+k)/dt
a= 8j
Answer:
(a) velocity, v = 8t j + k
(b) acceleration, a = 8 j
Explanation:
The position of the particle as a function of time is given as;
r = i + 4t² j + t k --------------------(i)
(a) To get the expression of its velocity, v, find the derivative of its position with respect to time by differentiating equation (i) with respect to t as follows;
v = dr / dt = 0 + 8t j + k
v = dr / dt = 8t j + k
v = 8t j + k ----------------------(ii)
Therefore, the equation/expression for the particle's velocity (v) is
v = 8t j + k
(b) To get the expression of its acceleration, a, find the derivative of its velocity with respect to time by differentiating equation (ii) with respect to t as follows;
a = dv / dt = t j + 0
a = dv / dt = t j
a = 8 j
Therefore, the expression for the particle's acceleration, a, is a = 8 j
A parallel-plate capacitor filled with air carries a charge Q. The battery is disconnected, and a slab of material with dielectric constant k = 2 is inserted between the plates. Which of the following statements is correct? The voltage across the capacitor decreases by a factor of 2. The charge on the plates is doubled. The voltage across the capacitor is doubled. The charge on the plates decreases by a factor of 2. The electric field is doubled.
Answer:
The voltage across the capacitor decreases by a factor of 2
Explanation:
When we put a dielectric material on a capacitor, we generate a polarization on it that changes the capacitance of the capacitor. The dielectric constant is the ratio between the initial capacitance (Co) without the dielectric and the new capacitance (C) with the dielectric:
[tex]k=\frac{C}{C_0} [/tex]
So, if k is equal to 2:
[tex] 2= \frac{C}{C_0}[/tex]
[tex]C=2*C_0 [/tex]
So, the capacitance is doubled with the dielectric
Because the battery is disconnected, the charge on the plates of capacitor must be constant because conservation of charge.
The effective electric field of a capacitor is the electric filed when we put a dielectric on it, and it is:
[tex]E_{eff}=\frac{E}{k}=\frac{E}{2} [/tex]
With E the initial electric field, so the electric field is halved
Finally, because we know electric field and voltage (V) on a parallel capacitor with distance between plates (d) are related by:
[tex] E=\frac{V}{d}[/tex]
Because electric field and voltage are directly proportional and d remains constant if Electric field is halved, then voltage is halved too. The voltage across the capacitor decreases by a factor of 2.
A spherical conductor has a radius of 14.0 cm and a charge of 42.0 µC. Calculate the electric field and the electric potential at the following distances from the center. (a) r = 8.0 cm magnitude direction electric field MN/C electric potential MV (b) r = 30.0 cm magnitude direction electric field MN/C electric potential MV (c) r = 14.0 cm magnitude direction electric field MN/C electric potential MV
Answer:
0 MN/C, 2.697 MV
4.1953 MN/C, 1.2586 MV
19.2642857143 MN/C, 2.697 MV
Explanation:
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
Electric field at r = 8 cm
E = 0 (inside)
Electric potential is given by
[tex]V=\dfrac{kq}{R}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV[/tex]
Electric potential is 2.697 MV
r = 30 cm
[tex]E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3^2}\\\Rightarrow E=4195333.33\ MN/C=4.1953\ MN/C[/tex]
Electric field is 4.1953 MN/C
[tex]V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3}\\\Rightarrow V=1258600\ V=1.2586\ MV[/tex]
Electric potential is 1.2586 MV
r = R = 14 cm
[tex]E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14^2}\\\Rightarrow E=19264285.7143\ N/C=19.2642857143\ MN/C[/tex]
The electric field is 19.2642857143 MN/C
[tex]V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV[/tex]
The potential is 2.697 MV
To calculate the electric field and electric potential at different distances from the center of a spherical conductor, use the equations for electric field and electric potential due to a point charge. At a distance of 8.0 cm, the electric field is 7.87 * 10^3 N/C and the electric potential is 5.25 kV. At a distance of 30.0 cm, the electric field is 1.26 * 10^3 N/C, but the electric potential cannot be calculated without a reference point. At 14.0 cm, the electric field and electric potential are zero.
Explanation:To calculate the electric field and electric potential at different distances from the center of a spherical conductor with a radius of 14.0 cm and a charge of 42.0 µC, we can use the equations for electric field and electric potential due to a point charge.
(a) At a distance of 8.0 cm from the center, the magnitude of the electric field can be calculated using the equation:
E = k * (Q/r²)
Substituting the given values, we get:
E = (9 * 10^9 Nm²/C²) * (42 * 10^-6 C) / (0.08 m)² = 7.87 * 10^3 N/C
To calculate the electric potential at this distance, we can use the equation:
V = k * (Q/r)
Substituting the given values, we get:
V = (9 * 10^9 Nm²/C²) * (42 * 10^-6 C) / (0.08 m) = 5.25 kV
(b) At a distance of 30.0 cm from the center, the magnitude of the electric field can be calculated in the same way:
E = (9 * 10^9 Nm²/C²) * (42 * 10^-6 C) / (0.3 m)² = 1.26 * 10^3 N/C
But the electric potential at this distance cannot be calculated using the given information, as we need the reference point for potential measurement.
(c) At a distance of 14.0 cm (the radius of the conductor) from the center, the electric field and electric potential are zero, as the conductor is in electrostatic equilibrium.
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If the temperature at the surface of Earth (at sea level) is 100°F, what is the temperature at 2000 feet if the average lapse rate is 3.5°F/1000 feet?
To find the temperature at 2000 feet above sea level, we need to use the average lapse rate of 3.5°F per 1000 feet. By calculating the temperature decrease with increasing altitude, we can determine that the temperature at 2000 feet would be 93°F.
Explanation:To find the temperature at 2000 feet, we need to use the average lapse rate. The average lapse rate tells us how much the temperature decreases with increasing altitude. In this case, the average lapse rate is 3.5°F per 1000 feet. So for every 1000 feet increase in altitude, the temperature decreases by 3.5°F.
Since we want to find the temperature at 2000 feet, which is 2000 feet above sea level, we need to divide 2000 by 1000 to find how many intervals of 1000 feet we have. There are 2 intervals. So, we multiply the lapse rate of 3.5°F by 2 to get a temperature decrease of 7°F.
Therefore, the temperature at 2000 feet would be 100°F - 7°F = 93°F.
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If the temperature at the surface of Earth (at sea level) is 100°F, at 2000 feet if the average lapse rate is 3.5°F/1000 the temperature is 93°F.
If the temperature at the surface of Earth at sea level is 100°F, we can calculate the temperature at 2000 feet using the average lapse rate, which is given as 3.5°F per 1000 feet.
The temperature change is 3.5°F/1000 feet × 2000 feet = 7°F.
Therefore, the temperature at 2000 feet above sea level would be the surface temperature minus the temperature change due to the lapse rate:
100°F - 7°F
= 93°F.
In what ways do the mirrors in X-ray telescopes differ from those found in optical instruments?
Explanation:
X-ray telescopes have a different design than that of optical telescopes, since x-rays can reflect off from mirror if they are struck at a particular of angle of the grazing. The X-rays are concentrated to a point in two reflections. By nesting the mirrors inside a one another , the X-ray telescope area can be enhanced.
Final answer:
X-ray telescope mirrors are designed to reflect high-energy X-rays at small angles, using precision-coated and aligned mirrors, while optical telescope mirrors only require front surface polishing and are used for visible light reflection, allowing them to be larger and more cost-effective.
Explanation:
The mirrors in X-ray telescopes, such as those used in the Chandra X-ray Observatory, are designed to reflect high-energy X-rays that are absorbed when incident perpendicular to the medium. Unlike mirrors in optical instruments, which reflect visible light usually by incident at direct angles, X-ray telescope mirrors reflect X-rays at small grazing angles, similar to a rock skipping across a lake. The design of these mirrors involves a long barrelled pathway with multiple pairs of mirrors, often coated with metals like iridium, to focus the rays at a specific point. They're precision-engineered to be extremely smooth for the most effective reflection.
In contrast, optical telescopes like reflectors use mirrors that only need the front surface polished accurately, avoiding issues like flaws and bubbles within the glass. These mirrors can be larger and more cost-effective than lenses, making them suitable for studying dimmer or more distant objects. Today's largest optical telescopes are reflectors for this reason.